C++第十二周【任务4】类的组合与继承

/*
* 程序的版权和版本声明部分
* Copyright (c) 2011, 烟台大学计算机学院学生
* All rights reserved.
* 文件名称：C++第十二周【任务4】                             
* 作    者：   李洪悬                             
* 完成日期：   2012  年  5  月  8  日

* 对任务及求解方法的描述部分
* 输入描述：圆的基本数据

* 问题描述：类的组合与继承

* 程序输出：圆的各种信息

*/

【任务4】类的组合与继承（1）先建立一个Point(点)类，包含数据成员x,y(坐标点)；（2）以Point为基类，派生出一个Circle(圆)类，增加数据成员(半径)，基类的成员表示圆心；（3）编写上述两类中的构造、析构函数及必要的输入输出函数（4）定义友元函数int locate，判断点p在圆c上、圆c内或圆c外，返回值<0圆内，==0圆上，>0 圆外；（5）重载关系运算符（6种）运算符，使之能够按圆的面积比较两个圆的大小；（6）给定一点p，求出该点与圆心相连成的直线与圆的两个交点并输出

BB平台的代码：

//用下面的main()函数测试

int main( )
{
Circle c1(3,2,4),c2(4,5,5);      //c2应该大于c1
Point p1(1,1),p2(3,-2),p3(7,3);  //分别位于c1内、上、外

cout<<"圆c1: "<<c1;
cout<<"点p1: "<<p1;
cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl;
cout<<"点p2: "<<p2;
cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl;
cout<<"点p3: "<<p3;
cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl;
cout<<endl;

cout<<"圆c1: "<<c1;
if(c1>c2) cout<<"大于"<<endl;
if(c1<c2) cout<<"小于"<<endl;
if(c1>=c2) cout<<"大于等于"<<endl;
if(c1<=c2) cout<<"小于等于"<<endl;
if(c1==c2) cout<<"等于"<<endl;
if(c1!=c2) cout<<"不等于"<<endl;
cout<<"圆c2: "<<c1;
cout<<endl;

Point p4,p5;
crossover_point1(p1,c1, p4, p5);

cout<<"点p1: "<<p1;
cout<<"与圆c1: "<<c1;
cout<<"的圆心相连，与圆交于两点，分别是："<<endl;
cout<<"交点: "<<p4;
cout<<"交点: "<<p5;
cout<<endl;

system("pause");
return 0;
}

 

程序源代码：

#include<iostream>

#include<Cmath>

using namespace std;

const double pi=3.14;

class Point //定义坐标点类
{
public:
Point()
{
x=0;

y=0;
}
Point(double x0,double y0)
{
x=x0;

y=y0;
}
double get_x()
{
return x;
}
double get_y()
{
return y;
}
void set_x(double n)
{
x=n;
}
void set_y(double n)
{
y=n;
}
friend ostream &operator << (ostream & output, Point & c);
private:

double x,y;   //点的横坐标和纵坐标
};
ostream &operator << (ostream & output, Point & c)
{
output<<"该点的横坐标为："<<c.x<<"    "<<"纵坐标为："<<c.y<<endl;

return output;
}
class Circle: public Point   //利用坐标点类定义圆类, 其基类的数据成员表示圆的中心
{
public:
Circle(double xx,double yy,double dd): Point(xx,yy) ,d(dd){}//构造函数

friend ostream &operator << (ostream & output, Circle & c);

friend double locate(Point &,Circle &);

friend void crossover_point1(Point &p1,Circle &c1,Point &p4,Point &p5);

bool operator > (Circle &t);
bool operator < (Circle &t);
bool operator >= (Circle &t);
bool operator <= (Circle &t);
bool operator == (Circle &t);
bool operator != (Circle &t);
private:

double d;
};
void crossover_point1(Point &p1,Circle &c1,Point &p4,Point &p5)
{
double n;
n=c1.get_x()+(sqrt((p1.get_x()-c1.get_x())*(p1.get_x()-c1.get_x ()))*c1.d )/(sqrt((p1.get_x ()-c1.get_x ())*(p1.get_x ()-c1.get_x ())+((p1.get_y ()-c1.get_y ())*(p1.get_y ()-c1.get_y ()))));
p4.set_x (n);
n=c1.get_x ()-(sqrt((p1.get_x ()-c1.get_x ())*(p1.get_x ()-c1.get_x ()))*c1.d )/(sqrt((p1.get_x ()-c1.get_x ())*(p1.get_x ()-c1.get_x ())+((p1.get_y ()-c1.get_y ())*(p1.get_y ()-c1.get_y ()))));
p5.set_x (n);
n=c1.get_y ()+(sqrt((p1.get_y ()-c1.get_y ())*(p1.get_y ()-c1.get_y ()))*c1.d )/(sqrt((p1.get_x ()-c1.get_x ())*(p1.get_x ()-c1.get_x ())+((p1.get_y ()-c1.get_y ())*(p1.get_y ()-c1.get_y ()))));
p4.set_y (n);
n=c1.get_y ()-(sqrt((p1.get_y ()-c1.get_y ())*(p1.get_y ()-c1.get_y ()))*c1.d )/(sqrt((p1.get_x ()-c1.get_x ())*(p1.get_x ()-c1.get_x ())+((p1.get_y ()-c1.get_y ())*(p1.get_y ()-c1.get_y ()))));
p5.set_y (n);
}
ostream &operator << (ostream & output, Circle & c)
{
output<<"圆的半径为："<<c.d<<"圆的圆心为"<<"("<<c.get_x()<<","<<c.get_y()<<")"<<endl;
return output;
}
double locate(Point &p,Circle &c)
{
double s,d,m;
s=(c.get_x()-p.get_x () )*(c.get_x()-p.get_x () )+(c.get_y ()-p.get_y () )*(c.get_y ()-p.get_y () );
m=sqrt(s);
d=m-c.d ;
return d;
}
bool Circle::operator > (Circle &t)
{
double s1,s2;
s1=pi*d*d;
s2=pi*t.d*t.d;
if(s1>s2 )
return true;
else
return false;
}
bool Circle::operator < (Circle &t)
{
double s1,s2;
s1=pi*d*d;
s2=pi*t.d*t.d;
if(s1<s2 )
return true;
else
return false;
}
bool Circle::operator >= (Circle &t)
{
double s1,s2;
s1=pi*d*d;
s2=pi*t.d*t.d;
if (s1<s2)
return false;
return true;
}

bool Circle::operator <= (Circle &t)
{
double s1,s2;
s1=pi*d*d;
s2=pi*t.d*t.d;
if (s1>s2)
return false;
return true;
}
bool Circle::operator == (Circle &t)
{
double s1,s2;
s1=pi*d*d;
s2=pi*t.d*t.d;
if (s1<s2)
return false;
if (s1>s2)
return false;
return false;
}

bool Circle::operator != (Circle &t)
{
double s1,s2;
s1=pi*d*d;
s2=pi*t.d*t.d;
if (s1==s2)
return false;
return true;
}
int main( )
{
Circle c1(3,2,4),c2(4,5,5);      //c2应该大于c1
Point p1(1,1),p2(3,-2),p3(7,3);  //分别位于c1内、上、外

cout<<"圆c1: "<<c1;
cout<<"点p1: "<<p1;
cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl;
cout<<"点p2: "<<p2;
cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl;
cout<<"点p3: "<<p3;
cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl;
cout<<endl;

cout<<"圆c1: "<<c1;
if(c1>c2) cout<<"大于"<<endl;
if(c1<c2) cout<<"小于"<<endl;
if(c1>=c2) cout<<"大于等于"<<endl;
if(c1<=c2) cout<<"小于等于"<<endl;
if(c1==c2) cout<<"等于"<<endl;
if(c1!=c2) cout<<"不等于"<<endl;
cout<<"圆c2: "<<c1;
cout<<endl;

Point p4,p5;
crossover_point1(p1,c1, p4, p5);

cout<<"点p1: "<<p1;
cout<<"与圆c1: "<<c1;
cout<<"的圆心相连，与圆交于两点，分别是："<<endl;
cout<<"交点: "<<p4;
cout<<"交点: "<<p5;
cout<<endl;

system("pause");
return 0;
}

运行结果：

圆c1: 圆的半径为：4圆的圆心为(3,2)

点p1: 该点的横坐标为：1    纵坐标为：1

点p1在圆c1之内

点p2: 该点的横坐标为：3    纵坐标为：-2

点p2在圆c1之上

点p3: 该点的横坐标为：7    纵坐标为：3

点p3在圆c1之外

圆c1: 圆的半径为：4圆的圆心为(3,2)

小于

小于等于

不等于

圆c2: 圆的半径为：4圆的圆心为(3,2)

点p1: 该点的横坐标为：1    纵坐标为：1

与圆c1: 圆的半径为：4圆的圆心为(3,2)

的圆心相连，与圆交于两点，分别是：

交点: 该点的横坐标为：6.57771    纵坐标为：3.78885

交点: 该点的横坐标为：-0.577709    纵坐标为：0.211146

请按任意键继续. . .

经验积累：理清思路，自顶向下，逐步求精！